Question

the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. if the tower is 50m high. find the distance between building and tower.

Answer 1

Step-by-step explanation:

Given height of tower CD=50m

Let the height of the building, AB=h

In right angled △BDC,

⇒  tan60o=BDCD

⇒  3=BD50

⇒  BD=350m

In right angled △ABD,

⇒  tan30o=BDAB

⇒  31=350h

∴  31=503h

∴  h=350=16.66m

∴  The height of the building is 16.66m.

Answer 2

Answer:

The angle of elevation of the top of a building from the foot of the tower is 30°.

The angle of elevation of the top of tower from the foot of the building is 60°.

Height of the tower is 50 m.

Let the height of the building be h.

Step-by-step explanation:

$\sf In \: \Delta BDC$

$:\implies \sf tan \: \theta = \dfrac{Perpendicular}{Base}$

$:\implies \sf tan \: 60^{\circ} = \dfrac{CD}{BD}$

$:\implies \sf \sqrt{3} = \dfrac{50}{BD}$

$:\implies \sf BD = \dfrac{50}{ \sqrt{3} } \: m$

___________________

$\sf In \: \Delta ABD,$

$\dashrightarrow\:\: \sf tan \: 30 ^{\circ} = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ \dfrac{50}{ \sqrt{3} } }$

$\dashrightarrow\:\: \sf h = \dfrac{ 50 }{ \sqrt{3} \times \sqrt{3} }$

$\dashrightarrow\:\: \underline{ \boxed{\sf Height= \dfrac{50}{3} \: m }}$

$\therefore\:\underline{\textsf{Height of the building is \textbf{50/3 meter}}}.$